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An increment of a Brownian motion between deterministic times $s<t$ is $W_t-W_s$. It has normal distribution $N(0,t-s)$ and is independent of the filtration up to time $s$. Increments on disjoint time intervals are independent. Their covariance is the length of overlap of the corresponding intervals; this also follows from the Brownian covariance kernel.

Standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ on a probability space with three defining properties: $B_0=0$ almost surely; its sample paths $t\mapsto B_t$ are continuous almost surely; and its Brownian increments over any finite collection of disjoint time intervals are independent, withHere $N(m,v)$ denotes the normal distribution with expectation $m$ and variance $v$. The alternative symbol $W_t$ denotes the same kind of stochastic process. Continuity is a simultaneous statement about an entire sample path, rather than a separate assertion at each fixed time. Relative to its natural Brownian filtration, a future Brownian increment is independent of all observations up to its starting time.

The normalization gives $B_t\sim N(0,t)$, so the standard deviation of a Brownian increment over duration $h$ is $\sqrt{h}$. For $h=1$, $0.01$, and $0.0001$, the corresponding standard deviations are $1$, $0.1$, and $0.01$. Time becoming one hundred times shorter makes typical displacement ten times smaller. Brownian scaling expresses this rule for the whole stochastic process, while the Brownian covariance kernel describes how positions at different times remain related despite independent increments.

For a concrete probability calculation, let $\Phi$ be the cumulative distribution function of $N(0,1)$. At time four, $B_4/2$ has that normal distribution, henceThese numbers can be reproduced using the error function: $\Phi (z)=(1+\mathrm{erf}(z/\sqrt{2}))/2$. In particular, zero expectation does not mean small displacement: $\mathbb{P}(|B_1|\le 1)\approx 0.6826894921$, but the same fixed interval contains only about $38.3\%$ of the probability at time four.

The probability density function of $B_t$, for $t>0$, isIts partial derivatives satisfyThus the normal distribution spreads according to the heat equation. Starting at a fixed position $x_0$ replaces $x$ by $x-x_0$. This is a useful bridge between individual random trajectories and the deterministic evolution of their probability density function: the same Brownian motion provides both descriptions.

Physical diffusion uses its own scale. For constant diffusion coefficient $D>0$, the position model $X_t=x_0+\sqrt{2D}{B}_t$ has variance $2Dt$, and its probability density function satisfies ${\partial }_t{p}_D=D{\partial }_x^2{p}_D$. For example, $D=0.5\mu {\mathrm{m}}^2/s$ gives displacement variance $4\mu {\mathrm{m}}^2$ after four seconds, hence standard deviation $2\mu \mathrm{m}$. Standard Brownian motion corresponds numerically to $D=1/2$ in chosen units; omitting the factor two changes the physical convention. Einstein's 1905 paper relates diffusion to the mean squared displacement in precisely this convention. ^[ref]

Continuity does not supply a smooth physical velocity. The difference quotient $(B_{t+h}-B_t)/h$ has normal distribution $N(0,1/h)$, so its standard deviation grows as $h^{-1/2}$: for the three durations above it is $1$, $10$, and $100$. This calculation motivates the nowhere differentiability of Brownian motion, but it alone is not a proof of the simultaneous assertion at every time. The theorem says that, with probability one, no finite derivative exists anywhere along the sample path. A drawn polygon only displays a finite grid; its straight segments cannot establish this theorem. See Theorem 7.1.6 in Probability: Theory and Examples (fifth edition) by Rick Durrett.

For physical particles, Brownian motion is a model of position on suitable observation scales. It does not claim that microscopic velocity is a derivative of this idealized path. Einstein explicitly restricts his assumption of independent increments to intervals that are not too short. ^[ref]

Another way to measure the roughness is quadratic variation. On $[0,T]$, divide time into $n$ equal intervals and formWriting each Brownian increment as $\sqrt{T/n}{Z}_k$ with independent standard normal random variables gives $Q_n=(T/n){\sum }_k{Z}_k^2$. Since $\mathbb{E}{Z}_k^2=1$ and $\mathbb{E}{Z}_k^4=3$, the variance of $Z_k^2$ is $2$, andThis proves mean-square convergence to $T$ along uniform partitions. For $T=2$ and $n=100$, the expectation is $2$, the variance is $0.08$, and the root mean square of the error is $\sqrt{0.08}\approx 0.2828427125$. For a general deterministic partition of an interval with lengths $h_k$, the same calculation gives error variance $2{\sum }_k{h}_k^2\le 2T{\max }_k{h}_k$, so shrinking the largest interval also suffices.

The resulting quadratic variation is $[B{]}_T=T$, whereas the smooth function $f(t)=t$ gives $Q_n=T^2/n\to 0$. This difference explains why the ordinary rules of calculus need the second-order correction in the Itô formula. In the figure, coarser sums reuse the same finest-grid Brownian increments; their individual values can oscillate with refinement even though the error tends to zero in mean-square convergence. Refining a fixed interpolating polygon forever would instead give zero quadratic variation, so the finite-grid illustration must not be confused with the limiting Brownian motion.

Historically, Brown's 1828 report describes microscopic observations made in 1827, starting with particles released from pollen and extending to inorganic materials; this challenged an explanation based solely on living matter. His observations and the later mathematical model are distinct contributions. ^[ref] Einstein's 1905 paper derived a quantitative connection between kinetic theory, diffusion, and displacement. ^[ref] Wiener's 1923 Differential-Space developed integration on spaces of paths, a foundation for Wiener measure; it also acknowledges earlier work and his own previous papers. These dates mark documented contributions rather than an assertion that each subject had a single discoverer. ^[ref]

The Brownian motion $B$ in the Dambis-Dubins-Schwarz theorem representation $M_t=B_{[M{]}_t}$ need not be independent of the bracket clock. Let $W$ be standard Brownian motion and takeThe Gaussian fourth moment and the Brownian covariance kernel give $\mathbb{E}[(W_t^2-t)W_s^2]=2s^2$ for $s\le t$, and henceIf the entire driving Brownian motion $B$ were independent of the clock $A$, conditioning on $A_t$ would give $\mathbb{E}[B_{A_t}{A}_t]=0$, contradicting this identity. Gaussian increments in bracket time therefore do not justify an independent random-clock model.

The displayed continuous centered Gaussian process, with value zero at time zero, has the Brownian covariance kernel. Hence it is a Brownian motion in its own natural filtration. More generally, replacing three by $c$ gives covariance $s+c(c-3)(s/2-s^2/(6t))$ for $0<s\le t$, so three is the only nonzero valid coefficient. Its conditional future increment in the original Brownian filtration is $3(t-s)t^{-1}(s^{-1}{\int }_0^s{W}_{u}du-W_s)$, and is generally nonzero.

Let $L$ be the triangular matrix with $L_{ij}=1$ for $j\le i$ and zero otherwise. Then $(L{L}^T{)}_{ij}=\min (i,j)$ and $\det L=1$, so the determinant of $L{L}^T$ is one. More generally, for $0=t_0<t_1<\cdots <t_n$, the matrix with entries $\min (t_i,t_j)$ factors as $L\mathrm{diag}(t_k-t_{k-1})L^T$ and has determinant ${\prod }_k(t_k-t_{k-1})$. This is the finite-dimensional structure behind the Brownian covariance kernel.

A spin echo refocuses a nucleus's fixed detuning. A nucleus that moves through a magnetic field gradient can have different detunings during the two intervals. To see the distinction without assuming that all noise is reversible, let $\Delta (t)$ vary during free evolution. With the same pulse convention, its residual phase at $2\tau$ isStatic detuning cancels because $\int y(t)dt=0$. A detuning ${\Delta }_a$ during the first interval and ${\Delta }_b$ during the second leaves $\Phi =\tau ({\Delta }_a-{\Delta }_b)$, which need not vanish. The final component signal is $e^{i\Phi }$ before relaxation; changing the overall receiver convention changes its conjugation but not its magnitude.

For an original quantitative model, take a constant magnetic field gradient $G=d{B}_z/dx$ and unrestricted one-dimensional Brownian motion $x(t)=x_0+\sqrt{2D}{W}_t$, where $D$ is the diffusion coefficient and $W_t$ is standard Brownian motion. Then $\Delta (t)={\Delta }_0+\gamma Gx(t)$. The constant terms cancel, and the Gaussian distribution of the remaining phase follows from the Brownian covariance kernel $\mathbb{E}[W_s{W}_t]=\min (s,t)$. Writing $Q(r)={\int }_r^{2\tau }y(t)dt$ givesThe characteristic function of a zero-mean Gaussian distribution now supplies the extra factor $\mathbb{E}[e^{i\Phi }]=e^{-\mathrm{Var}\Phi /2}$, soHere $\gamma$ is in radians per second per tesla, $G$ in tesla per metre, and $D$ in square metres per second: the extra exponent is dimensionless. The figure chooses $\gamma /(2\pi )=42.58MHz/T$, $G=0.010T/m$ and $D=2.0\times 10^{-9}{\mathrm{m}}^2/s$. With the earlier $\tau$ and $T_2$, diffusion reduces the echo from $0.60653$ to about $0.32930$. These are illustrative model parameters, not values extracted from Hahn's experiment.

This result assumes instantaneous perfect pulses, a constant magnetic field gradient, unrestricted diffusion with constant diffusion coefficient, and independent homogeneous transverse relaxation. Restricted motion, changing gradients, finite pulses and interactions between spins can require different attenuation laws. Even a fixed Hamiltonian is refocused only for the terms whose sign the pulse reverses: a simultaneous 180° rotation of two spins leaves their $J_{z,1}{J}_{z,2}$ product unchanged, because both factors change sign. A recovered echo therefore demonstrates a particular recoverable phase history, rather than reversal of every microscopic process.

A centered Gaussian process with almost surely continuous sample paths and covariance $\mathbb{E}[B_s{B}_t]=\min (s,t)$ is standard Brownian motion. Here Gaussian process means that every finite-dimensional distribution is a multivariate normal distribution; having separate normal distributions at each time would not suffice. This characterization is stated in Pitman, Basic Properties of Brownian Motion, Definition 15.2.

To compute the Brownian covariance kernel directly, suppose $0\le s\le t$. Since $B_t=B_s+(B_t-B_s)$ and the added Brownian increment is centered and independent of $B_s$,Symmetry gives $\min (s,t)$ for arbitrary nonnegative times. For instance, $\mathrm{Cov}(B_1,B_4)=1$ and their correlation is $1/\sqrt{1\cdot 4}=1/2$. Positions are therefore related even though disjoint Brownian increments are independent: both positions contain the displacement accumulated during the first unit of time. Conversely, taking differences of $\min (s,t)$ gives zero covariance for disjoint increments; their joint multivariate normal distribution turns this into independence.

Take the real L2 space $\mathcal{H}=L^2((0,\infty ),dt)$ and the feature mapEach indicator function belongs to $\mathcal{H}$ because its squared L2 norm is $x<\infty$. Their inner product is the length of the overlapping intervals:This realizes $k_1$ as a positive-semidefinite kernel; it is also the Brownian covariance kernel.

For $0\le s<t$, linearity of the isonormal Gaussian process givesThe squared $L^2$ norm of this indicator is $t-s$, so part (a) yieldsAlso, the covariance formula gives $\mathrm{Cov}(W_s,W_t)=s\wedge t$, the Brownian covariance kernel. The Gaussian statement concerns the signed increment.

In the real Hilbert space $L^2({\mathbb{R}}_{+})$, let $f_t=1_{[0,t]}$. Then $k_i(t)=⟨f_t,h_i⟩$. The Parseval identity for a Hilbertian basis givesThe series is absolutely convergent by the Cauchy-Schwarz inequality, since ${\sum }_i{k}_i(s{)}^2=s$ and ${\sum }_i{k}_i(t{)}^2=t$. Thusthe Brownian covariance kernel. This is the Brownian covariance from an integrated orthonormal basis.

For $t>0$, write $A_t=t^{-1}{\int }_0^t{W}_{u}du$. Deterministic integrals of the Gaussian process $W$ are Gaussian, as follows by taking $L^2$ limits of Riemann sums, so $(W,A)$ is jointly Gaussian. For $0<s\le t$, integrating the Brownian covariance kernel givesFor the last equality, integrate ${\int }_0^t\min (u,v)dv=ut-u^2/2$ over $0\le u\le s$ and divide by $st$. ConsequentlyThe only nonzero choice giving Brownian covariance is $\boxed{c=3}$; even the variance condition at $s=t$ requires $c(c-3)=0$.

Define ${\hat{W}}_0=0$. Since $|A_t|\le {\sup }_{u\le t}|W_u|\to 0$ almost surely as $t\downarrow 0$, the transformed process is continuous at zero, as well as at positive times. It is centered Gaussian and has covariance $s\wedge t$, so part (c) provesThe transformed process is Brownian in its own natural filtration. This is the Brownian motion transform by three times its running average. The filtration qualification matters: for $t>s>0$,which is not identically zero. The Gaussian covariance calculation identifies the Brownian filtration generated by $\hat{W}$; it does not make the transformed process a martingale in the larger original Brownian filtration.

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.

This is the Brownian covariance kernel, so its integral operator is a covariance operator. To obtain its eigendecomposition, suppose $Uf=\lambda f$ with $\lambda \ne 0$. Splitting the integral at $s$ givesDifferentiation yields $\lambda f^{\prime }(s)={\int }_s^{1}f(t)dt$ and $\lambda f^{\prime \prime }(s)=-f(s)$, with boundary conditions $f(0)=0$ and $f^{\prime }(1)=0$. Hence the normalized eigenpairs areThe eigenvalues are positive and summable, consistently with positivity and the trace-class operator property of a covariance operator.

Standard Brownian motion is centered, so $m_t=0$. If $s\le t$, write $B_t=B_s+(B_t-B_s)$; the final increment is independent of $B_s$ and has mean zero. ThusThis is the Brownian covariance kernel. Together with joint Gaussianity and continuous paths, it gives the Gaussian-process characterization of Brownian motion.